Is there a way to round numbers into a reader friendly format? (e.g. $1.1k) [closed]

Question

Much like the Stackoverlow reputation rounding, I'm hoping to do the same thing with currency

$1,000 => 1k

$1,000,000 => 1m

How can I achieve this in JavaScript (preferably in jQuery)?

Flutter

This currently doesn't qualify as an acceptable code-golf question. see meta.stackexchange.com/questions/24258 You need a few more test cases, to make it community wiki and finally you should accept the shortest (in code size) answer. Single language code-golfs are boring, you should also open it to other languages – Lissotrichous

David created a [code-golf] version: #2692823 – Caffrey

Thanks guys -- as you can see I'm new to Stack Overflow. I'm excited that you all think this is an interesting problem! – Flutter

Answer 1

Here is a simple function to do it:

function abbrNum(number, decPlaces) {
  // 2 decimal places => 100, 3 => 1000, etc
  decPlaces = Math.pow(10, decPlaces);

  // Enumerate number abbreviations
  var abbrev = ["k", "m", "b", "t"];

  // Go through the array backwards, so we do the largest first
  for (var i = abbrev.length - 1; i >= 0; i--) {

    // Convert array index to "1000", "1000000", etc
    var size = Math.pow(10, (i + 1) * 3);

    // If the number is bigger or equal do the abbreviation
    if (size <= number) {
      // Here, we multiply by decPlaces, round, and then divide by decPlaces.
      // This gives us nice rounding to a particular decimal place.
      number = Math.round(number * decPlaces / size) / decPlaces;

      // Handle special case where we round up to the next abbreviation
      if ((number == 1000) && (i < abbrev.length - 1)) {
        number = 1;
        i++;
      }

      // Add the letter for the abbreviation
      number += abbrev[i];

      // We are done... stop
      break;
    }
  }

  return number;
}
console.log(abbrNum(1200, 3))

Outputs:

abbrNum(12 , 1)          => 12
abbrNum(0 , 2)           => 0
abbrNum(1234 , 0)        => 1k
abbrNum(34567 , 2)       => 34.57k
abbrNum(918395 , 1)      => 918.4k
abbrNum(2134124 , 2)     => 2.13m
abbrNum(47475782130 , 2) => 47.48b

Demo: http://jsfiddle.net/jtbowden/SbqKL/

Warehouseman

this feels golfy. how small can we get this function? – Insistency

FUND IT. Seriously though, I'd love to see that in a golf thread. – Astigmia

@ Damien Wilson: Code-golf tag added. What do you mean by fund? – Flutter

@Baloney: I was just making it flexible. Do what is best for your application. An "appropriate cut-off" point is often subjective and/or application specific. – Warehouseman

Code golf version posted here. For contributing members of OP question: feel free to change the parameters of the question...I had a difficult time deciding the limitations/boundaries. – Insistency

function a(n,d){x=(''+n).length,p=Math.pow,d=p(10,d);x-=x%3;return Math.round(n*d/p(10,x))/d+" kMGTPE"[x/3]} --- Sufficiently Golfed (108)? :) – Laurasia

abbrNum(999950, 0) => 1000k, yeah it only happens 50 / 1M times but still. – Luffa

@crizCraig: Good catch! I added a special case handler for that in the code. There is probably a more efficient way to do it, but it works for now. – Warehouseman

104 bytes function(a,b,c,d){c=(''+a).length;d=Math.pow;b=d(10,b);return((a*b/d(10,c-=c%3))+.5|0)/b+' kMGTPE'[c/3]} – Prig

https://mcmap.net/q/181926/-abbreviate-a-localized-number-in-javascript-for-thousands-1k-and-millions-1m – Sparker

I have an optimised version of the above code in objective C over here. https://mcmap.net/q/183091/-ios-convert-large-numbers-to-smaller-format. ... .... @JeffB : your method doesn't work for negative numbers. you need to use the following ..... if(size <= Math.abs(Math.round(number))) ..... instead of ..... if(size <= number) ......... Which has an extra benefit of removing your code block of 'Handle special case where we round up to the next abbreviation' – Relevance

can without round ? abbrNum(1432, 1); => 1.4k; abbrNum(1992, 1); => 1.9k – Matos

@Matos Just change the Math.round above to Math.floor. – Warehouseman

Answer 2

var floor=Math.floor, abs=Math.abs, log=Math.log, round=Math.round, min=Math.min;
var abbrev = ['k', 'Mil', 'Bil']; // abbreviations in steps of 1000x; extensible if need to edit

function rnd(n, precision) {
    var prec = 10**precision;
    return round(n*prec)/prec;
}

function format(n) {
    var base = floor(log(abs(n))/log(1000));
    var suffix = abbrev[min(abbrev.length-1, base-1)];
    base = abbrev.indexOf(suffix) + 1;
    return suffix ? rnd(n/1000**base,2)+suffix : ''+n;
}

Demo:

> tests = [-1001, -1, 0, 1, 2.5, 999, 1234, 
           1234.5, 1000001, 10**9, 10**12]
> tests.forEach(function(x){ console.log(x,format(x)) })

-1001 "-1k"
-1 "-1"
0 "0"
1 "1"
2.5 "2.5"
999 "999"
1234 "1.23k"
1234.5 "1.23k"
1000001 "1Mil"
1000000000 "1Bil"
1000000000000 "1000Bil"

Billiebilling

This one is the best since it works with negative numbers too – Santanasantayana

@shrekuu: seems to be passed just fine as rounding to two decimal places in round(n/pow(...), 2); feel free to clarify – Billiebilling

Oh sorry my bad. Failed to read it more carefully. I will delete my previous comment to avoid misleading. – Aftergrowth